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Let's solve the problem step by step. We are given the equation of a curve \( C \) in the form:

\[
y = 2x^2 + px + q
\]
where \( p \) and \( q \) are integers. We also know that the curve has a stationary point at \( (3, -5) \).

### Part (a)
We are asked to show that \( p = -12 \) and to find the value of \( q \).

#### Step 1: Differentiate the equation of the curve

To find the stationary points, we need to compute the derivative of \( y \) with respect to \( x \). So, let's differentiate:

\[
\frac{dy}{dx} = 4x + p
\]

At a stationary point, the derivative is zero. We know the stationary point occurs at \( x = 3 \), so substitute \( x = 3 \) into the derivative:

\[
4(3) + p = 0
\]
\[
12 + p = 0
\]
\[
p = -12
\]

This confirms that \( p = -12 \).

#### Step 2: Find \( q \)

We are also given that the curve passes through the point \( (3, -5) \). Substitute \( x = 3 \), \( y = -5 \), and \( p = -12 \) into the equation of the curve:

\[
-5 = 2(3)^2 + (-12)(3) + q
\]
\[
-5 = 18 - 36 + q
\]
\[
-5 = -18 + q
\]
\[
q = 13
\]

Thus, the value of \( q \) is \( 13 \).

### Part (b)
We are asked to state, giving a reason, the nature of the stationary point.

#### Step 1: Second derivative test

To determine the nature of the stationary point, we use the second derivative. Differentiate \( \frac{dy}{dx} = 4x - 12 \) again with respect to \( x \):

\[
\frac{d^2y}{dx^2} = 4
\]

Since \( \frac{d^2y}{dx^2} = 4 \), which is positive, the stationary point at \( (3, -5) \) is a **minimum**.

### Part (c)
We are asked to find the equation of the normal to the curve at the point where \( x = 1 \).

#### Step 1: Find the gradient of the tangent at \( x = 1 \)

The gradient of the tangent to the curve is given by the first derivative, \( \frac{dy}{dx} = 4x - 12 \). Substituting \( x = 1 \):

\[
\frac{dy}{dx} = 4(1) - 12 = 4 - 12 = -8
\]

The gradient of the tangent is \( -8 \), so the gradient of the normal is the negative reciprocal of this:

\[
\text{Gradient of the normal} = \frac{1}{8}
\]

#### Step 2: Find the coordinates of the point on the curve where \( x = 1 \)

Substitute \( x = 1 \) into the equation of the curve to find the \( y \)-coordinate:

\[
y = 2(1)^2 - 12(1) + 13
\]
\[
y = 2 - 12 + 13 = 3
\]

So, the point on the curve is \( (1, 3) \).

#### Step 3: Equation of the normal

The equation of a line is given by:

\[
y - y_1 = m(x - x_1)
\]

where \( m \) is the gradient of the normal, and \( (x_1, y_1) \) is the point \( (1, 3) \). Substituting the values:

\[
y - 3 = \frac{1}{8}(x - 1)
\]

Multiply through by 8 to eliminate the fraction:

\[
8(y - 3) = x - 1
\]
\[
8y - 24 = x - 1
\]
\[
x - 8y + 23 = 0
\]

Thus, the equation of the normal is:

\[
x - 8y + 23 = 0
\]

---

### Summary of answers:
- (a) \( p = -12 \), \( q = 13 \)
- (b) The stationary point is a minimum because \( \frac{d^2y}{dx^2} = 4 \), which is positive.
- (c) The equation of the normal is \( x - 8y + 23 = 0 \).

---

Do you have any questions or need further details?

### Related Questions:
1. How do you find the second derivative of more complex functions?
2. What does the second derivative tell us about a curve's shape?
3. Can there be stationary points that are neither maxima nor minima?
4. How do we find the tangent to a curve at a given point?
5. What role does the discriminant play in determining the nature of stationary points?

**Tip:** Always check if the second derivative is positive or negative at a stationary point to confirm whether it is a minimum or maximum.

The curve C has equation y = 2x^2 + px + q where p and q are integers. The curve C has a stationary point at (3, -5). (a) Show that p = -12 and find the value of q. (b) State, giving a reason, the nature of the stationary point. (c) Find an equation of the normal to C at the point on C where x = 1. Give your answer in the form ax + by + c = 0.

Learn how to find stationary points and equations of normals for the quadratic curve y = 2x^2 + px + q, where p = -12 and q = 13. This problem involves differentiation, the second derivative test, and calculating normals. Suitable for students in Grades 10-12.

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